Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - 1}{x + 10} = \dfrac{-8x + 19}{x + 10}$
Explanation: Multiply both sides by $x + 10$ $ \dfrac{x^2 - 1}{x + 10} (x + 10) = \dfrac{-8x + 19}{x + 10} (x + 10)$ $ x^2 - 1 = -8x + 19$ Subtract $-8x + 19$ from both sides: $ x^2 - 1 - (-8x + 19) = -8x + 19 - (-8x + 19)$ $ x^2 - 1 + 8x - 19 = 0$ $ x^2 - 20 + 8x = 0$ Factor the expression: $ (x - 2)(x + 10) = 0$ Therefore $x = 2$ or $x = -10$ However, the original expression is undefined when $x = -10$. Therefore, the only solution is $x = 2$.